Question: What is the extraneous solution to these equations? $\dfrac{x^2 - 3x}{x + 5} = \dfrac{-13x - 21}{x + 5}$
Answer: Multiply both sides by $x + 5$ $ \dfrac{x^2 - 3x}{x + 5} (x + 5) = \dfrac{-13x - 21}{x + 5} (x + 5)$ $ x^2 - 3x = -13x - 21$ Subtract $-13x - 21$ from both sides: $ x^2 - 3x - (-13x - 21) = -13x - 21 - (-13x - 21)$ $ x^2 - 3x + 13x + 21 = 0$ $ x^2 + 10x + 21 = 0$ Factor the expression: $ (x + 7)(x + 3) = 0$ Therefore $x = -7$ or $x = -3$ The original expression is defined at $x = -7$ and $x = -3$, so there are no extraneous solutions.